let f(z) = (z/(z^2 −z+2)) developp f at integr serie.


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Question Number 39038 by maxmathsup by imad last updated on 01/Jul/18

$l e t f (z) = \frac{z}{z^{2} - z + 2}$ $d e v e l o p p f a t i n t e g r s e r i e .$
Commented by prof Abdo imad last updated on 02/Jul/18
$let decompose f(x) ondide C(x) Δ=1−4(2)=−7=(i(√7))^2 ⇒ z_1 =((1+i(√7))/2) and z_2 =((1−i(√7))/2) ⇒ f(z) = (z/((z−z_1 )(z−z_2 ))) = (a/(z−z_1 )) +(b/(z−z_2 )) a= (z_1 /(z_1 −z_2 )) = (z_1 /(i(√7))) =((−iz_1 )/(√7)) b = (z_2 /(z_2 −z_1 )) = (z_2 /(−i(√7))) = ((iz_2 )/(√7)) ⇒ f(x)= ((−iz_1 )/((√7)(z−z_1 ))) +((iz_2 )/((√7)(z−z_2 ))) ⇒ f^((n)) (x) =((−iz_1 )/(√7)){ (((−1)^n n!)/((z−z_1 )^(n+1) ))} +((iz_2 )/(√7)){ (((−1)^n n!)/((z−z_2 )^(n+1) ))} = (((−1)^n n!)/(√7)){ (z_2 /((z−z_2 )^(n+1) )) −(z_1 /((z−z_1 )^(n+1) ))} ⇒ f^((n)) (0)=(((−1)^n n!)/(√7)){ (z_2 /((−1)^(n+1) z_2 ^(n+1) ))−(z_1 /((−1)^(n+1) z_1 ^(n+1) ))} = ((n!)/(√7)){ ((−z_2 )/z_2 ^(n+1) ) +(z_1 /z_1 ^(n+1) )} finally f(z) =Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) z^n =Σ_(n=0) ^∞ (1/(√7))( (1/z_1 ^n ) −(1/z_2 ^n ))z^n$
$l e t d e c o m p o s e f (x) o n d i d e C (x)$ $Δ = 1 - 4 (2) = - 7 = {(i \sqrt{7})}^{2} \Rightarrow$ $z_{1} = \frac{1 + i \sqrt{7}}{2} a n d z_{2} = \frac{1 - i \sqrt{7}}{2} \Rightarrow$ $f (z) = \frac{z}{(z - z_{1}) (z - z_{2})} = \frac{a}{z - z_{1}} + \frac{b}{z - z_{2}}$ $a = \frac{z_{1}}{z_{1} - z_{2}} = \frac{z_{1}}{i \sqrt{7}} = \frac{- {i z}_{1}}{\sqrt{7}}$ $b = \frac{z_{2}}{z_{2} - z_{1}} = \frac{z_{2}}{- i \sqrt{7}} = \frac{{i z}_{2}}{\sqrt{7}} \Rightarrow$ $f (x) = \frac{- {i z}_{1}}{\sqrt{7} (z - z_{1})} + \frac{{i z}_{2}}{\sqrt{7} (z - z_{2})} \Rightarrow$ $f^{(n)} (x) = \frac{- {i z}_{1}}{\sqrt{7}} {\frac{{(- 1)}^{n} n!}{{(z - z_{1})}^{n + 1}}} + \frac{{i z}_{2}}{\sqrt{7}} {\frac{{(- 1)}^{n} n!}{{(z - z_{2})}^{n + 1}}}$ $= \frac{{(- 1)}^{n} n!}{\sqrt{7}} {\frac{z_{2}}{{(z - z_{2})}^{n + 1}} - \frac{z_{1}}{{(z - z_{1})}^{n + 1}}} \Rightarrow$ $f^{(n)} (0) = \frac{{(- 1)}^{n} n!}{\sqrt{7}} {\frac{z_{2}}{{(- 1)}^{n + 1} z_{2}^{n + 1}} - \frac{z_{1}}{{(- 1)}^{n + 1} z_{1}^{n + 1}}}$ $= \frac{n!}{\sqrt{7}} {\frac{- z_{2}}{z_{2}^{n + 1}} + \frac{z_{1}}{z_{1}^{n + 1}}} f i n a l l y$ $f (z) = \sum_{n = 0}^{\infty} \frac{f^{(n)} (0)}{n!} z^{n} = \sum_{n = 0}^{\infty} \frac{1}{\sqrt{7}} (\frac{1}{z_{1}^{n}} - \frac{1}{z_{2}^{n}}) z^{n}$
Commented by prof Abdo imad last updated on 02/Jul/18
$let simplify (1/z_1 ^n ) −(1/z_2 ^n ) = ((z_2 ^n −z_1 ^n )/((z_1 z_2 )^n )) = (((((1−i(√7))/2))^n −( ((1+i(√7))/2))^n )/2) = −(1/2^(n+1) ) { (1+i(√7))^n −(1−i(√7))^n } =−(1/2^(n+1) ) { Σ_(k=0) ^n C_n ^k (i(√7))^k −Σ_(k=0) ^n C_n ^k (−i(√7))^k } =((−1)/2^(n+1) ) Σ_(k=0) ^n C_n ^k { (i(√7))^k −(−i(√7))^k } =((−1)/2^(n+1) ) Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) { 2i(−1)^p 7^p (√7) } =((−i)/2^n ) (√7)Σ_(p=0) ^([((n−1)/2)]) C_n ^(2p+1) (−7)^p ⇒ f(z)=Σ_(n=0) ^∞ ((−i)/2^n ){ Σ_(p=0) ^([((n−1)/2)]) (−7)^(p ) C_n ^(2p+1) }z^n$
$l e t s i m p l i f y \frac{1}{z_{1}^{n}} - \frac{1}{z_{2}^{n}}$ $= \frac{z_{2}^{n} - z_{1}^{n}}{{(z_{1} z_{2})}^{n}} = \frac{{(\frac{1 - i \sqrt{7}}{2})}^{n} - {(\frac{1 + i \sqrt{7}}{2})}^{n}}{2}$ $= - \frac{1}{2^{n + 1}} {{(1 + i \sqrt{7})}^{n} - {(1 - i \sqrt{7})}^{n}}$ $= - \frac{1}{2^{n + 1}} {\sum_{k = 0}^{n} C_{n}^{k} {(i \sqrt{7})}^{k} - \sum_{k = 0}^{n} C_{n}^{k} {(- i \sqrt{7})}^{k}}$ $= \frac{- 1}{2^{n + 1}} \sum_{k = 0}^{n} C_{n}^{k} {{(i \sqrt{7})}^{k} - {(- i \sqrt{7})}^{k}}$ $= \frac{- 1}{2^{n + 1}} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {2 i {(- 1)}^{p} 7^{p} \sqrt{7}}$ $= \frac{- i}{2^{n}} \sqrt{7} \sum_{p = 0}^{[\frac{n - 1}{2}]} C_{n}^{2 p + 1} {(- 7)}^{p} \Rightarrow$ $f (z) = \sum_{n = 0}^{\infty} \frac{- i}{2^{n}} {\sum_{p = 0}^{[\frac{n - 1}{2}]} {(- 7)}^{p} C_{n}^{2 p + 1}} z^{n}$
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